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Q.
The locus of the point of intersection of the lines $(\sqrt{3}) kx + ky -4 \sqrt{3}=0$ and
$\sqrt{3} x-y-4(\sqrt{3}) k=0$ is a conic, whose eccentricity is
$K =\frac{4 \sqrt{3}}{\sqrt{3} x + y }=\frac{\sqrt{3} x - y }{4 \sqrt{3}}$
$\Rightarrow 3 x ^{2}- y ^{2}=48$
$\Rightarrow \frac{ x ^{2}}{16}-\frac{ y ^{2}}{48}=1$
Now, $48=16\left( e ^{2}-1\right)$
$\Rightarrow e =\sqrt{4}=2$